Update: The contest ranking is similar to ACM ICPC style and penalty for wrong submission has been set to 10 minutes.

Less than half hour to go for the contest.

Update for Codechef team: Top 5 Global and Top 5 Indians get 250 laddus.

Contest has started. Best of luck to everyone.

You can find the intended solutions here. I will be uploading the short editorials in a while. Detailed ones will be put up on codechef discuss forum by tomorrow. Hope you all enjoyed the contest.

Any feedback regarding the clarity of problems, time limits, test data being weak/strong etc?

Short Editorial of first 4 problems:

Problem 1 — Matrix Game Greedy solution applies and Matrix has no rolw in the problem. Just consider the numbers and at each moment of time every person tries to take the largest available number. Find the sum they can accumulate after this process and print answer accordingly.

Problem 2 — Stingy Strings Again greedy solution applies in the sense that we replace all occurences of a given character by its corresponding number or replace none of them. Next part is to modify the cost function as "length — 2*(count of numbers) + (sum of numbers)/k". Using this, we can say that we replace only those ccharacters with numbers whose value is greater than or equal to 2*k.

Problem 3 — Maze love

Let number of steps taken in each direction be N, S, E, W. So we have following equations:

N — S = m, E — W = n, Na + Sb + Ec + Wd = P

Replace S, W in last equation, we get N * A + E * B = C, where A = a + b, B = c + d, C = P + mb + nd

The other constraints are N, S, E, W are integers and N >= m, S >= 0 and E >= n, W>= 0.

Solution exists when gcd(A, B) divides C. After this check, we simply brute force over the possible values of N and update answer (i.e. minimum value of N+S+E+W) if present.

Problem 4 — Replace and Substring queries.

Idea is to find the number of substrings with given mask in string B easily. For this, we fix a particular index and find the first point of occurrence of every character on the right and add the corresponding contribution to the mask. For example- In "aabbbbdac" for index 1, "a" occurs at "1", "b" occurs at "3", "d" occurs at "7" and "c" occurs at "9". So number of substring with one endpoint at 1 and having mask 1 are 2, mask 3 are 3, mask 11 are 2 and mask 15 is 1.

One this initial computation is done, we try to handle the queries and updates. For query on string "a", we just need to find the mask of the substring a[l:r] and this requires to compute if a particular character occurs in a range. With updates on string "a", this can be easily handled using fenwick tree or segment tree.

To handle updates on string "b", idea is to first remove contribution of every substring which contains the xth character and then add the contribution back. For finding number of mask of substrings with contain xth character, we see the substring can either end at x, start at x or contain x in the middle. The first 2 are easy to calculate and are similar to the construction by finding first occurrence to the left and right of index x. For the substring which contain "x" in the middle we can visualise them as 2 parts, left and right and their mask being the "OR" of the left and right mask. Thus doing this OR convolution of the left and right masks in naive manner i.e. O(ALPHA**2) we can find their contribution too.

Complexity of precomputation is O(m * ALPHA * log m) and query and update on string "a" is O(ALPHA * logn) and O(log n) while update on string "b" is O(ALPHA * log m + ALPHA**2).

Is the expected solution of Raid Systems O(log(y) * log(n)) per query of type 2? I barely got AC using this.

Short explanation for last problem: Raid Systems

The first idea of the problem is to see that files on xth hard disk can be retrieved using files on hard disk 1. The file indices differ by exactly (x-1). For example, if at any moment of time, hard disk 1 has files f1, f2, f5 when n = 5, then hard disk 2 has files f(1+2-1), f(3+2-1), f(5+2-1) i.e. f2, and f3 (f7 is discarding). See the table to convince regarding this observation. Thus, the queries of both type can be handled very easily using binary search only provided we are able to find the files in sorted order on hard disk 1 on any day.

Next thing is to notice is that the pattern of files being stored on hard disk 1 will repeat after some time, to be precise after 2**(ceil(log2(n)). Draw a table for n = 8 and n = 5 to get view of it. Next observation is that total number of files we require for all possible days is bounded by 3**ceil(log2(n)). For example: Consider n = 4

0: f1

1: f1, f2, f3, f4

2: f1, f3

3: f1, f2

4: same as 0 i.e. pattern repeats.

The total number of files stored i.e. slots used is (1 + 4 + 2 + 2) = 9, 3**2. You can see this thing for any general n. Complete the table for n=5 and observe the same. Now, we just need to come up with a 3**(ceil(log2(n)) generation algorithm, which gives us the files in sorted order on any day. The memory complexity will also be the same. Many constructive algorithms can exist and you can try to find your own. My solution uses just one of my finding, idea is again similar to binary search i.e. can be divide the files (in sorted order) in half, calculate the files on left side and use it to calculate the files on right side. Basically, I mean to say that if a particular bit is set in "y" and we calculate the lower half files, then the upper half files are basically the mirror image on the lower half files. You can check my "gen" function in the solution code for details. Code

Good Problemset. I particularly liked the problem "Replace and Substring queries"

Sorry for the delays, the editorials are given in below links:

1. ICL1801
2. ICL1802
3. ICL1803
4. ICL1804
5. ICL1805